Submission #348106

#	Time	Username	Problem	Language	Result	Execution time	Memory
348106		Kerim	Rectangles (IOI19_rect)	C++17	37 / 100	1737 ms	458128 KiB

rect

#include "rect.h"
#include "bits/stdc++.h"
#define MAXN 100009
#define INF 1000000007
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(__typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
#define debug(x,y,z)  cerr<< #x <<" = "<< x<<" "<< #y <<" = "<< y<<" "<< #z <<" = "<< z<<endl;
using namespace std;
 
typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
const int N=2505;
int ata[N],mn[N],mx[N],arr[N],c[N],sz,sub[N],vis[N],rc,pos;
vector<int>row[N][N],col[N][N];
vector<pair<PII,int> >addc[N];
vector<PII>remc[N];
int F[N][N];
bool cmp(int x,int y){
	return (arr[x]<arr[y]);	
}
int tap(int x){
	if(ata[x]==x)
		return x;
	return ata[x]=tap(ata[x]);	
}
void merge(int x,int y){
	if((x=tap(x))==(y=tap(y)))
		return;
	if(sub[x]<sub[y])
		swap(x,y);
	ata[y]=x;
	sub[x]+=sub[y];
	umin(mn[x],mn[y]);
	umax(mx[x],mx[y]);	
}
map<pair<int,PII>,int>pm[2];
//The key idea is there are at most 2*n^2 such pairs
void add(int x,int y){
	assert(x+1<y);
	//if rc=0, then rectangle on row (pos) in range (arr[pos][x],arr[pos][y]) is good
	//if rc=1, then rectangle on column (pos) in range (arr[x][pos],arr[y][pos]) is good
	if(pm[rc][mp(pos,mp(x,y))])
		return;
	pm[rc][mp(pos,mp(x,y))]=1;	
	if(!rc)
		row[x][y].pb(pos);
	else
		col[x][y].pb(pos);
}
void solve(int x){
	if(x>1 and vis[x-1]){
		int a=mn[tap(x-1)]-1;
		if(a>=1)
			add(a,x);
	}	
	if(x<sz and vis[x+1]){
		int b=mx[tap(x+1)]+1;
		if(b<=sz)
			add(x,b);
	}
}
void upd(int x){
	vis[x]=1;
	if(x>1 and vis[x-1])
		merge(x-1,x);
	if(x<sz and vis[x+1])
		merge(x,x+1);	
}
void f(){
	for(int i=1;i<=sz;i++)
    	ata[i]=mn[i]=mx[i]=i,sub[i]=1,c[i]=i,vis[i]=0;
    sort(c+1,c+sz+1,cmp);
    for(int i=1;i<=sz;i++){
    	int j=i;
		while(j<=sz and arr[c[j]]==arr[c[i]])
			solve(c[j++]);
		for(int k=i;k<j;k++)
			upd(c[k]);
		i=j-1;
    }
}
void upd(int y,int x,int val){
	for(;x<MAXN;x+=x&(-x))
		F[y][x]+=val;	
}
int get(int y,int x){
	int res=0;
	for(;x;x-=x&(-x))
		res+=F[y][x];
	return res;
}	
long long count_rectangles(std::vector<std::vector<int> > a) {
	int n=int(a.size()),m=int(a[0].size());
	if(n<3 or m<3)
		return 0;
	{
		sz=m;
		for(int j=1;j<=n;j++){
			for(int i=1;i<=m;i++)
				arr[i]=a[j-1][i-1];
			rc=0;pos=j;f();
		}sz=n;
		for(int j=1;j<=m;j++){
			for(int i=1;i<=n;i++)
				arr[i]=a[i-1][j-1];
			rc=1;pos=j;f();
		}
	}
	int tmp,st=-1,en,cnt,who;
	for(int i=1;i<=n;i++)	
		for(int j=i+2;j<=n;j++){
			tmp=int(col[i][j].size());
			for(int k=0;k<tmp;k++){
				if(st==-1)st=col[i][j][k];
				en=col[i][j][k];
				if(k+1==tmp or col[i][j][k]+1!=col[i][j][k+1]){
				//	printf("%d %d ~ %d %d\n",i,j,st,en);
					for(int h=st;h<=en;h++)
						addc[h].pb(mp(mp(i+1,j-1),en));
					//remc[en].pb(mp(i+1,j-1));
					st=-1;
				}
			}
		}
	ll ans=0;
	for(int i=2;i<m;i++){
		tr(it,addc[i]){
			upd(it->ff.ss,it->ff.ff,+1);
		//	printf("inc %d ~ %d %d\n",i,it->ff.ff,it->ff.ss);
			//F[it->ff.ss][it->ff.ff]++;
			remc[it->ss].pb(it->ff);
		}
		for(int j=i;j<=m;j++){
			if(j<m){
				tmp=int(row[i-1][j+1].size());cnt=0;
				for(int k=0;k<tmp;k++){who=row[i-1][j+1][k];
					//printf("%d %d %d\n",i-1,j+1,who);
					cnt++;ans+=get(who,who)-get(who,who-cnt);
					//for(int h=who-cnt;h<=who;h++)ans+=F[who][h];
					if(k+1==tmp or who+1!=row[i-1][j+1][k+1])cnt=0;
				}
				//wr
			}
			tr(it,remc[j]){
				upd(it->ss,it->ff,-1);
				//F[it->ss][it->ff]--;
				//printf("dec %d ~ %d %d\n",j,it->ff,it->ss);
			}remc[j].clear();
		}
	}
	return ans;
}

• Subtask 1: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 12 / 12
• Subtask 4: 0 / 22
• Subtask 5: 10 / 10
• Subtask 6: 0 / 13
• Subtask 7: 0 / 28